Solving Trigonometric Equations Exercise 4
Exercises
Solution: \(x=0,\pi,\frac{3\pi}{2}, 2\pi\)
Solution Method: While it might be tempting to divide everything by a \(\tan^2{x}\) here, if we do we’ll be losing a solution to the equation! (Think about \(x^2=3x\). We can clearly see that \(x=0\) is a solution to the equation. But if we divide by x, we may think \(x=3\) is the only solution.) So we cannot divide trig values out of the equation, but we can factor them out. For this equation, we have a common factor of \(\tan^2{x}\), so I can factor it out of the LHS. \[\begin{aligned} \tan^2{x}+\tan^2{x}\sin{x}&=0 \\ \tan^2{x}(1+\sin{x}) &= 0 \end{aligned}\] So I can set \(\tan^2{x}\) and \(1+\sin{x}\) equal to 0 separately.
\[\begin{aligned} \tan^2{x} &= 0 \\ \sqrt{\tan^2{x}} &= \sqrt{0} \\ \tan{x} &= 0 \\ \end{aligned}\] I know that \(\tan{x}\) is defined by the ratio identity \(\tan{x} = \frac{\sin{x}}{\cos{x}}\). So for \(\tan{x}=0\), \(\sin{x}\) must be 0. Cosine equaling zero will make tangent undefined, NOT zero! And \(\sin{x}=0\) is satisfied by the quadrantal angles \(x=0,\pi,2\pi\) (We must include both 0 and \(2\pi\) because the interval includes both as endpoints).
\[\begin{aligned} 1+\sin{x} &= 0 \\ \sin{x} &= -1 \end{aligned}\] Thinking back to the unit circle, I know that \(\sin{x} = -1\) is only satisfied in \([0,2\pi]\) by the quadrantal angle \(x=\frac{3\pi}{2}\).
So the values in \([0,2\pi]\) satisfying \(\tan^2{x}+\tan^2{x}\sin{x}=0\) are \(x=0,\pi,\frac{3\pi}{2}, 2\pi.\)
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